• ArbitraryValue@sh.itjust.works
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    8 hours ago

    You don’t need calculus to do this. Neither one is accelerating, so “5 seconds after they started moving” is irrelevant. Just calculate the velocity of one in the reference frame of the other by subtracting the vectors: from the point of view of the boy, the girl’s velocity vector has orthogonal components of -5 ft/sec north and 1 ft/sec east, so the magnitude is 26^0.5 ft/sec.

    • andros_rex@lemmy.world
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      59 minutes ago

      I mean, there’s quite a few calculus problems that you can do without calculus. Pretty much 80% of the optimization problems/max’s and minimums in a typical Calc 1 class can be done if you remember that -b/2a is the vertex of a quadratic.

    • Mirodir@discuss.tchncs.de
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      6 hours ago

      The “5 seconds after they started moving” is relevant. If we assume this takes place on Earth (i.e. on the surface of a sphere with a set pair of north/south poles), the angle between the two vectors changes depending on their current position.

      If it’s not on the equator, it’s also slightly up to interpretation if “Due East” means they’ll turn to stay on the same latitude, always adjusting to stay moving east forever or if they’ll do a great circle. In the former case, the north moving one will eventually get stuck at the north-pole too instead of continuing their circle around the globe. Most likely not within 5 seconds though, unless the place they started was within 25 feet of the north-pole.

      To actually do the math we’ll need to know (or somehow deduce) where “the place where everything about them began” is though.

    • rustydrd@sh.itjust.works
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      6 hours ago

      I guess the calculus portion of this is to write the separation as a function of time, s = √26*t, and then realize that the rate of separation is the same regardless of time, because the first derivative is a constant.

  • ooterness@lemmy.world
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    8 hours ago

    At any given time T, the coordinates form a right triangle with legs of length 5T and T. Therefore the distance D is given by D^2 = (5T)^2 + T^2 = 26T^2. This simplifies to D = T * sqrt(26). Therefore the rate of separation is sqrt(26) ft/sec regardless of time

  • farmgineer@nord.pub
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    9 hours ago

    Shouldn’t it be ‘after having been together’?

    What is ‘at the same time’ referring to in that sentence? They wanted to break up at the same time (as in both had the idea)? They wanted to break up at the same time on the clock to continue the theme of things being same-y?

    The boy is due north of what? The place? The girl? Also, the girl should be wondering about her decision, I think.

    (I don’t even speak English every day anymore, so I could be wrong).

    • Bonsoir@lemmy.ca
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      9 hours ago

      They said goodbye at a given position and are then leaving each in a different direction. They start to move at the same time from the same point.

      • NoSpotOfGround@lemmy.world
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        8 hours ago

        Wait, we know their position exactly? That means we have no idea what their velocities are!

        Actually, their velocities are specified precisely in the problem description.

        What? Velocity too? Now we know nothing!!

    • misericordiae@literature.cafe
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      7 hours ago

      (I don’t even speak English every day anymore, so I could be wrong).

      You’re not wrong. I think some of it is the difference between casual speech and formal writing (people are more likely to say “after being” but write “after having been”, especially in published work)**, but some of it is also just poorly phrased. It makes enough sense to a native speaker to get what the problem is asking, though.

      __

      ** I think the first may be correct in some cases, but idk the rule.

    • bridgeburner@lemmy.world
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      6 hours ago

      Read th question again. It’s not about how far apart they are after 5 seconds. It’s about how fast they are separating from each other, so it’s about a rate, not distance.